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Question Number 139918 by mathsuji last updated on 02/May/21

$$1+cos2°+cos4°+cos6°+...+cos88°$$

Answered by mr W last updated on 02/May/21

$$\underset{k=0}{\sum ^{44}}(\cos \frac{2k\pi }{180}+i\sin \frac{2k\pi }{180})=\underset{k=0}{\sum ^{44}}{e}^{\frac{2k\pi }{180}i}$$$$=\frac{1-e^{\frac{2\pi i}{180}\times 45}}{1-e^{\frac{2\pi i}{180}}}=\frac{1-i}{(1-\cos \frac{\pi }{90})-i\sin \frac{\pi }{90}}$$$$=\frac{(1-i)[(1-\cos \frac{\pi }{90})+i\sin \frac{\pi }{90}]}{{(1-\cos \frac{\pi }{90})}^2+{\sin }^2\frac{\pi }{90}}$$$$=\frac{1-\cos \frac{\pi }{90}+\sin \frac{\pi }{90}-i(1-\cos \frac{\pi }{90}+\sin \frac{\pi }{90})}{{(1-\cos \frac{\pi }{90})}^2+{\sin }^2\frac{\pi }{90}}$$$$=\frac{1-\cos \frac{\pi }{90}+\sin \frac{\pi }{90}-i(1-\cos \frac{\pi }{90}+\sin \frac{\pi }{90})}{2(1-\cos \frac{\pi }{90})}$$$$$$$$1+\cos 2°+\cos 4°+...+\cos 88°=$$$$=\underset{k=0}{\sum ^{44}}\cos \frac{2k\pi }{180}$$$$=\frac{1-\cos \frac{\pi }{90}+\sin \frac{\pi }{90}}{2(1-\cos \frac{\pi }{90})}$$$$=\frac{1}{2}(1+\cot \frac{\pi }{180})$$$$=\frac{1}{2}(1+\cot 1°)$$

Commented by mathsuji last updated on 02/May/21

$$thankyouverymuchSir..$$

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